Problem: The equation of a circle $C$ is $x^2+y^2+12x+6y+41 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2+12x) + (y^2+6y) = -41$ $(x^2+12x+36) + (y^2+6y+9) = -41 + 36 + 9$ $(x+6)^{2} + (y+3)^{2} = 4 = 2^2$ Thus, $(h, k) = (-6, -3)$ and $r = 2$.